# Calculating Max. Demand Kilovolt-Amperes and Required Cable Current Rating

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## A building site and electrical equipment

A building site is to have the following electrical equipment available for use:

• Tower crane, electric motors totalling 75 kW at 415 V;
• Sump pump, 5 kW at 240 V;
• 60 tungsten lamps of 100 W each at 240 V;
• 12 flood lamps of 400 W each at 240 V;
• 20 hand tools of 400 W at 110 V.

To calculate //

1. Find the total kilovolt-amperes to be supplied to the site if the power factor of all rotary equipment is 0.8.
2. Find the electrical current rating for the incoming supply cable to the site.
3. Estimate the cost of electricity consumed on the site during a 12-month contract.

### Ok, let’s do it…

For rotating machinery, formulae for power in Volt Amperes [VA] goes like this:

For single-phase current:

For three-phase current:

The results of the power calculations are given in Table 1 below. The answers required are as follows.

1. The total input power kilovolt-amperes required for site is 120.8 kVA

2. The incoming supply cable capacity at 415 volt, three-phase, 50 Hz required is:

This is the input current to the site at the voltage of that cable. This is not the same as a total of the currents calculated in Table 1 as these larger numbers only appear at their reduced voltages in the relevant sub-circuits.

3. Assume that the crane, pump and tools are running for 25% of an 8-h working day, 5 days per week for 48 weeks, 20 of the tungsten lamps are for security lighting 16 h every night, and the remaining 40 tungsten lamps and the flood lamps are used for 3 h per day, 5 days per week for the winter period of 20 weeks.

The crane, pump and tools are working for:

The security lamps are working for:

The other lamps are working for:

#### Table 1 – Building site plant schedule

 Equipment Power (kW) Number kW kVA V A Tower Crane 75 1 75 93.75 415 130.4 Sump Pump 5 1 5 6.25 240 26 Lamps 0.1 60 6 6 240 25 Flood Lamps 0.4 12 4.8 4.8 240 20 Hand Tools 0.4 20 8 10 110 90.9 Total // 98.8 120.8 n.a. n.a.

#### Table 2 – Building site energy use

 Equipment Power (kW) Number Time (h) Energy (kWh) Tower Crane 75 1 480 36000 Sump Pump 5 1 480 2400 Tungsten lamps, security Lamps 0.1 20 5824 11648 Lamps 0.1 40 300 1200 Flood Lamps 0.4 12 300 1440 Hand Tools 0.4 20 480 3840 Total Power Used // 56528

The total energy used by the systems is found from:

kWh = number of appliances × kW per appliance × operation hours as shown in Table 2.

If electricity costs 8 p per unit (kWh) then the estimated cost for the 1-year contract will be:

Reference: Building Services Engineering // Electrical Installations – David V. Chadderton (Purchase paperback from Amazon)

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#### Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry fascilities. Professional in AutoCAD programming. Present on

1. Rohit subramanian
Mar 03, 2017

I have problem in one of the site where the connecting load is High HP pump -150 and 60 with 500kva transformer connected .
the MD is 185 connected obtained but the ideal time is so high that power factor is getting low checked out the parameters additional 1.5KVAR given for the grid than also the power factor is not getting maintained.

last month MD shooted up to 193.4 KVA but the running of the plant was minimal .
Help out in this regards

2. ALI
Jan 18, 2017

hay i am doing analysis on base of MDI penalty can anyone please tell me that how it is calculated. in bill i find readings like this

PRV CUMM MDI 98 PRS CUMM MDI 104 MDI 6

what does this mean….

3. magdy
Aug 17, 2015

we just start a construction of abig project at ajyad of mekkah saudi arabia we wanna make calculation of the total power needed for service using demand and load factor please tel me me how to compute the demand factor

4. Munaf
Aug 15, 2015

Thank you for the information

5. K.Ruju
Mar 04, 2015

Thanks for the Information

6. jesus g
Feb 21, 2015

thanks

7. Loum Peter
Feb 08, 2015

I’m so sorry i had forgotten the principle of the power triangle so sorry really the 93.75kva for the crane is correct.

8. Bimal
Feb 08, 2015

Nice good and very helpful

9. Thomas
Feb 07, 2015

How about the power consumption for testing& commissioning?

10. Loum Peter
Feb 06, 2015

Thanks for your information, but one think..for three phase , power W=1.732xV(line)xI(line)xPF no? if so that means VA =W/(1.732×0.8) and the KVA for the crane would be 54.13KVA….

• jesus perez
Feb 07, 2015

No. Active power (W) is equal to VA x power factor

11. sandar
Feb 05, 2015

Thanks u for ur information

Feb 03, 2015

Thanks for the information.

13. Bonginkosi
Feb 03, 2015

Thank you Edvard Csanyi for an informative and summarized example.
it assisted me in a project I am doing now.
I keep reading your articles, these are very helpful articles
the whole page is very useful, keep up the good work